integral of 1/(1+\sqrt[4]{x-1)}

Lösung

\int \frac{1}{1 + 4 x - 1} d x

4 (\frac{( 1 + 4 x - 1 ) ^{3}}{3} - \frac{3 ( 1 + 4 x - 1 ) ^{2}}{2} + 3 (1 + 4 x - 1) - ln 1 + 4 x - 1) + C

Schritte zur Lösung

\int \frac{1}{1 + 4 x - 1} d x

Wende U-Substitution an

= \int \frac{4 u ^{3}}{1 + u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{u ^{3}}{1 + u} d u

Wende U-Substitution an

= 4 \cdot \int \frac{( v - 1 ) ^{3}}{v} d v

Multipliziere aus

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

= 4 \cdot \int v^{2} - 3 v + 3 - \frac{1}{v} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int v^{2} d v - \int 3 v d v + \int 3 d v - \int \frac{1}{v} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 3 v d v = \frac{3 v ^{2}}{2}

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 4 (\frac{v ^{3}}{3} - \frac{3 v ^{2}}{2} + 3 v - ln ∣ v ∣)

Ersetze zurück

= 4 (\frac{( 1 + 4 x - 1 ) ^{3}}{3} - \frac{3 ( 1 + 4 x - 1 ) ^{2}}{2} + 3 (1 + 4 x - 1) - ln 1 + 4 x - 1)

Füge eine Konstante zur Lösung hinzu

= 4 (\frac{( 1 + 4 x - 1 ) ^{3}}{3} - \frac{3 ( 1 + 4 x - 1 ) ^{2}}{2} + 3 (1 + 4 x - 1) - ln 1 + 4 x - 1) + C

t an g e n t y = \frac{5 x}{1 + x ^{2}}, (5, \frac{25}{26})

\int w csc (4 w^{2} - 1) d w

\frac{d}{d x} (1.5 x - 0.25 x^{2})

l a pl a ce t r an s f or m cos (x)

\frac{d}{d x} (x e^{- x})